Page 35 - Foundations Of Differential Calculus

18 1. On Finite Differences
we have Σaz = ay, provided that a is a constant. Since ∆x = ω,wehave
2
Σω = x+C and Σaω = ax+C; since ω is a constant, we have Σω = ωx+C,
3 2
Σω = ω + C, and so forth.
27. If we invert the differences of powers of x which we previously found,
we have Σω = x and from this Σ1 = x/ω. Then we have

2 2
Σ 2ωx + ω = x ,
so that
x 2 ω x 2 x
Σx = − Σ = − .
2ω 2 2ω 2
Furthermore,

2 2 3 3
Σ 3ωx +3ω x + ω = x ,
or
2 2 3 3
3ωΣx +3ω Σx + ω Σ1 = x ,
so that

2 x 3 ω 2
Σx = − ωΣx − Σ1,
3ω 3
and so
3 2
2 x x ωx
Σx = − + .
3ω 2 6
In a similar way we have

3 x 4 3ω 2 2 ω 3
Σx = − Σx − ω Σx − Σ1.
4ω 2 4
2
If for Σx ,Σx, and Σ1 we substitute the previously found values, we obtain
3 x 4 x 3 ωx 2
Σx = − + .
4ω 2 4
Then, since
5 4
4 x 3 2 3 ω
Σx = − 2ωΣx − 2ωΣx − ω Σx − Σ1,
5ω 5
when we make the appropriate substitutions we have

4 x 5 1 4 1 3 1 3
Σx = − x + ωx − ω x.
5ω 2 3 30